Section 2.1: Shift Ciphers and Modular Arithmetic

Practice HW from Barr Textbook

(not to hand in)

p.66 # 1, 2, 3-6, 9-12, 13, 15

Modular Arithmetic

• In grade school, we first learned how to divide numbers.

Example 1: Consider

Determine the quotient and remainder of this

division and write the result as an equation.

3403

40

Solution:

• The previous example illustrates a special case of the division algorithm which we state next.

• Before stating this algorithm, recall that the integers are the numbers in the following set:

Integers: },4,3,2,1,0,1,2,3,4{

Division algorithm

• Let m be a positive integer (m > 0) and let b be any integer. Then there is exactly one pair of integers q (called the quotient) and r (called the remainder) such that

where .rqmb mr 0

• A number of primary interest in this class will be the remainder that we obtain the division of two numbers. We will find the remainder so often that we use a special term that is used to describe its computation. This is done in the following definition.

Definition

• We say that r is equal to b MOD m , written as b MOD m , if r is the integer remainder of b divided by m . We define the variable m as the modulus.

Example 2: Determine 25 MOD 7, 31 MOD 5,

26 MOD 2, and 5 MOD 7.

Solution:

Note:

• In the division algorithm, the remainder r is non-negative, that is, . This fact means that when doing modular arithmetic that we will never obtain a negative remainder. To compute b MOD m when b < 0 correctly, we must always look for the largest number that m evenly divides that is less than b. The next example illustrates this fact.

0r

Example 3: Compare computing 23 MOD 9 with -23 MOD 9.

Solution:

Doing Modular Arithmetic For Larger Numbers With A Calculator

• To do modular arithmetic with a calculator, we use the fact from the division algorithm that

and solve for the remainder to obtain

rqmb

qmbr

• We put this result in division tableau format as follows:

q

rqmb

qm

bm

Truncated Quotient (chop digits to right of decimal)

Remainder

Example 4: Compute 1024 MOD 37:

Solution:

Example 5: Compute 500234 MOD 10301

Solution:

Example 6: Compute -3071 MOD 107

Solution:

Generalization of Modular Arithmetic• In number theory, modular arithmetic has

a more formal representation which we now give a brief description of. This idea can be expressed with the following example.

Example 7: Suppose we want to find a solution

to the equation b MOD 7 = 4

Solution:

• The numbers

from Example 7 that give a remainder of 4 MOD 7 represents a congruence class. We define this idea more precisely in the following definition.

},32,25,18,11,4,3,10,17,{

Definition

• Let m be a positive integer (the modulus of our arithmetic). Two integers a and b are said to be congruent modulo m if a-b is divisible by m. We write

(note the lower

case “mod”)mba mod

Note

• The previous definition can be thought of more informally as follows. We say that

if a and b give the same integer remainder when divided by m. That is,

if

r = a MOD m = b MOD m.

mba mod

mba mod

Example 8: Illustrate why .

Solution:

7mod1125

• The last example illustrates that when the uppercase MOD notation is used, we are interested in only the specific integer remainder when a number is divided by a modulus. The lowercase mod notation with the notation is used when we are looking for a set of numbers that have the same integer remainder when divided by a modulus. In this class, we will primarily use the MOD notation.

*Note:

• When considering b MOD m, since ,

the only possible remainders are

This causes the remainders to “wrap” around when performing modular arithmetic. This next example illustrates this idea.

mr 0

.1,,2,1,0 m

Example 9: Make a table of y values for the

equation

y = (x + 5) MOD 9

Solution:

Modular Equation Addition Prop.

• Solving equations (and congruences) if modular arithmetic is similar to solving equations in the real number system. That is, if

then

and

for any number k.

mba mod

mkbka mod

mkbka mod

Example 10: Make a list of five solutions to

mod 8

Solution:

27 x

Basic Concepts of Cryptography

• Cryptography is the art of transmitting information in a secret manner. We next describe some of the basic terminology and concepts we will use in this class involving cryptography.

• Plaintext – the actual undisguised message (usually an English message) that we want to send.

• Ciphertext – the secret disguised message that is transmitted.

• Encryption (encipherment) – the process of converting plaintext to ciphertext.

• Decryption (decipherment) – process of converting ciphertext back to plaintext.

Notation

• represents all possible remainders in a MOD system, that is,

• For representing our alphabet, we use a MOD 26 system

mZ

}1,2,,2,1,0{ mmZ m

}25,24,,2,1,0{26 Z

MOD 26 Alphabet Assignment

• Used to perform a one to one correspondence between the alphabet letters and the elements of this set.

Monoalphabetic Ciphers

• Monoalphabetic Ciphers are substitution ciphers in which the correspondents agree on a rearrangement (permutation) of the alphabet. In this class, we examine 3 basic types of monoalphabetic ciphers

Types of Monoalphabetic Ciphers

1. Shift Ciphers (covered in Section 2.1)

2. Affine Ciphers (covered in Section 2.2)

3. Substitution Ciphers (covered in Section 2.3)

Shift Ciphers• If x is a numerical plaintext letter, we encipher x

by computing the

Enciphering formula for Shift Ciphers

MOD 26, where k is in

Here y will be the numerical ciphertext letter.

*Note: k is called the key of the cipher and represents the shift amount.

)( kxy 26Z

Example 11: The Caesar cipher, developed byJulius Caesar

is a shift cipher given by

MOD 26

Note that the key k = 3.

)3( xy

MOD 26

Use the Caesar cipher to create a cipher

alphabet. Then use it to encipher the message

“RADFORD”.

Solution: To create the cipher alphabet, we

substitute the MOD 26 alphabet assignment

number for each letter into the Caesar shift

cipher formula and calculate the corresponding

ciphertext letter number as follows:

DMODMODyxA 326326)30(0

FMODMODyxC 526526)32(2

GMODMODyxD 626626)33(3

ZMODMODyxW 25262526)322(22

AMODMODyxX 0262626)323(23

BMODMODyxY 1262726)324(24

CMODMODyxZ 2262826)325(25

EMODMODyxB 426426)31(1

This gives the corresponding correspondence

between the plain and ciphertext alphabets

Using the above table, we can encipher the

message “RADFORD” as follows

Hence, the ciphertext is “UDGIRUG” . █

Plain A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Cipher D E F G H I J K L M N O P Q R S T U V W X Y Z A B C

Plaintext: R A D F O R D

Ciphertext: U D G I R U G

In the last example we did not have to create the

entire plain and ciphertext alphabets to encipher the message. We could instead just used the shift

cipher formula y = (x+3) MOD 26 directly.

The Caesar cipher is just a special case of a shift

cipher with a key of k = 3. In a general shift

cipher, the key k can be any value in a MOD 26 system, that is, any value in the set

}25,24,,2,1,0{

Example 12: Encipher the message “SEINFELD”

using a 12 shift cipher.

Solution:

Deciphering Shift Ciphers

• Given a key k, plaintext letter number x, and ciphertext letter number y, we decipher as follows:

y = (x + k) MOD 26

Deciphering formula for shift ciphers

x = (y – k) MOD 26

where y is the numerical ciphertext letter, x is the numerical plaintext letter, and k is the key of the cipher (the shift amount).

*Note: In the deciphering shift cipher formula,

– k MOD 26 can be converted to its equivalent positive form by finding a positive remainder.

Example 13: Suppose we received the ciphertext

“YLUJLQLD” that was encrypted using a Caesar

cipher (shift ). Decipher this message.

Solution:

Example 14: Decipher the message

“EVZCJZXDFE” that was enciphered using a 17

shift cipher.

Solution: Using the shift cipher formula

y = (x + k) MOD 26,

we see that the key for this cipher must be

k = 17. Hence the formula becomes

y = (x + 17) MOD 26

Recall in this formula that x represents the

alphabet assignment number for the plaintext and

y represents the alphabet assignment number for

the cipher-text. Since we want to decipher the

above cipher-text, we must solve the above

equation for x. Rearranging first gives:

x + 17 = y MOD 26

To solve for x, we must subtract 17 from both

sides. This gives:

x = (y – 17) MOD 26 (*)

Since – 17 MOD 26 = 9 (this can be computed

simply by taking –17 + 26 = 9), we can write

equation (*) as:

x = (y + 9) MOD 26 (**)

Either equations (*) or (**) can be used to

decipher the message. We will use equation (**).

Taking each letter of the cipher-text

“EVZCJZXDFE” and using the MOD 26 alphabet

assignment, we obtain:

NMODMODxyE 13261326)94(4

EMODMODxyV 4263026)921(21

IMODMODxyZ 8263426)925(25

LMODMODxyC 11261126)92(2

SMODMODxyJ 18261826)99(9

IMODMODxyZ 8263426)925(25

GMODMODxyX 6263226)923(23

MMODMODxyD 12261226)93(3

OMODMODxyF 14261426)95(5

NMODMODxyE 13261326)94(4

Hence, the plaintext is “NEIL SIGMON”. █

Cryptanalysis of Shift Ciphers • As the last two examples illustrate, one must

know the key k used in a shift cipher when deciphering a message. This leads to an important question. How can we decipher a message in a shift cipher if we do not know the key k ? Cryptanalysis is the process of trying to break a cipher by finding its key. Cryptanalysis in general is not an easy problem. The more secure a cipher is, the harder it is to cryptanalyze. We will soon see that a shift cipher is not very secure and is relatively easy to break.

Methods for Breaking a Shift Cipher

1. Knowing x = (y – k) MOD 26, we can test all possibilities for k (there are 26 total in a MOD 26 alphabet )

until we recover a message that makes sense.

}25,24,,2,1,0{

2. Frequency analysis: Uses the fact that the most frequently occurring letters in the ciphertext produced by shift cipher has a good chance of corresponding to the most frequently occurring letters in the standard English alphabet. The most frequently occurring letters in English are E, T, A, O, I, N, and T (see the English frequency table).

We will demonstrate these techniques using Maplets.